What is the enthalpy of formation of $NH_3$ if the bond enthalpies are $(N \equiv N) = 941 \ kJ/mol$,$(H-H) = 436 \ kJ/mol$,and $(N-H) = 389 \ kJ/mol$?

Which of the following equations corresponds to the definition of enthalpy of formation at $298 \ K$?

$C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)} \dots \dots(I) \quad \Delta H = -393 \, kJ \, mol^{-1}$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_{2}O_{(l)} \dots \dots(II) \quad \Delta H = -287.3 \, kJ \, mol^{-1}$
$2CO_{2(g)} + 3H_{2}O_{(l)}$ $\rightarrow C_{2}H_{5}OH_{(l)} + 3O_{2(g)} \dots \dots(III) \quad \Delta H = 1366.8 \, kJ \, mol^{-1}$
Find the standard enthalpy of formation of $C_{2}H_{5}OH_{(l)}$.

The heat of formation is the change in enthalpy accompanying the formation of a substance from its elements at $298 \ K$ and $1 \ atm$ pressure. Since the enthalpies of elements are taken to be zero,the heat of formation $(\Delta H_f)$ of compounds

If the value of $\Delta H$ in a reaction is positive,then the reaction is called

In the reaction for the transition of carbon in the diamond form to carbon in the graphite form,$\Delta H = -453.5 \ cal$. This points out that